[sameeralord]

Hello guys,

I have a question regarding the equilibrium constant. When the volume of the reacted vessel is doubled the concentration of all reactants and products are halved. Then according to Le chateliers principle the reaction moves to the more particle side. My question is why should that happen because all the actions of Le Chatelier principle is to maintain the equilibirum constant. If so if

K= P/R

Then both the products and reactant are multiplied by half and nothing would happen to K. So why do they have move to more particle side can't they just remain the same.

Any help would be appreciated. Thanks!!

[cliverlong]

The key is to consider an example where all reactants and products are in the gas phase and the are unequal number of moles on either side of the equation

consider

NO₂ in equilibrium with NO and O₂

1. Write a balanced equation for this reaction
2. Count number of moles on either side
3. Consider a scenario where the equilibrium is equal to the number of moles in the balanced equation. So if moles of balanced equation are in ratio 1 NO₂ :2NO :3 O₂ K_p will be p(products) / p(reactants) = (3 x 2) / 1
4. Now consider halving or doubling the volume as in your scenario.
5. Work out the new quotient if partial pressures do not change
6. Work out what needs to happen to the partial pressures of each component in order to restore the original K_p
7. Have the partial pressures stayed the same or changed?
8. What does that mean in terms of the ratio of the reactants and products? Same or changed?

Clive

[sameeralord]

The key is to consider an example where all reactants and products are in the gas phase and the are unequal number of moles on either side of the equation

consider

NO₂ in equilibrium with NO and O₂

1. Write a balanced equation for this reaction
2. Count number of moles on either side
3. Consider a scenario where the equilibrium is equal to the number of moles in the balanced equation. So if moles of balanced equation are in ratio 1 NO₂ :2NO :3 O₂ K_p will be p(products) / p(reactants) = (3 x 2) / 1
4. Now consider halving or doubling the volume as in your scenario.
5. Work out the new quotient if partial pressures do not change
6. Work out what needs to happen to the partial pressures of each component in order to restore the original K_p
7. Have the partial pressures stayed the same or changed?
8. What does that mean in terms of the ratio of the reactants and products? Same or changed?

Clive

Thanks for the reply Clivelong. .

2NO₂------ 2NO + O₂

kp = (2/5 x P)² X (1/5 x P)  / (2/5 x P)²

Assume P=1

hence constant=  1/5

If volume is doubled

kp = (2/5 x P)² X (1/5 x P)  / (2/5 x P)²

So = (4/25 x P²/4) X (1/5 x P/2)  / (4/25 x P²/4)

    = (1/10 x P)
 constant now is= 1/10 to maintain the constant you need less reactants

So if Pressure is halved the to maintain the constant you need more products hence it moves to more particle side.

So how about with dilutions where aqeuous solutions are included then you don't use Kp. How can we prove that one?

[cliverlong]

... to maintain the constant you need less reactants

So if Pressure is halved the to maintain the constant you need more products hence it moves to more particle side.

Looks good to me! Can you describe the result using the "language" of Le Chatelier?

So how about with dilutions where aqeuous solutions are included then you don't use Kp. How can we prove that one?

Have a look here

http://www.chemguide.co.uk/physical/equilibria/kc.html

(basically it is the same method you have already applied)


Clive